Difference between revisions of "2010 AMC 10A Problems/Problem 11"

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== Problem 11 ==
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The length of the interval of solutions of the inequality <math>a \le 2x + 3 \le b</math> is <math>10</math>. What is <math>b - a</math>?
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<math>
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\mathrm{(A)}\ 6
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\qquad
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\mathrm{(B)}\ 10
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\qquad
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\mathrm{(C)}\ 15
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\qquad
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\mathrm{(D)}\ 20
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\qquad
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\mathrm{(E)}\ 30
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</math>
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==Solution==
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Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>.  
 
Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>.  
  
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We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math>
 
We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math>
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== See also ==
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{{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}}

Revision as of 09:54, 2 January 2012

Problem 11

The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$. What is $b - a$?

$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$


Solution

Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$.

$a\le 2x+3\le b$

Subtract $3$ from all of the quantities:

$a-3\le 2x\le b-3$

Divide all of the quantities by $2$.

$\frac{a-3}{2}\le x\le \frac{b-3}{2}$

Since we have the range of the solutions, we can make them equal to $10$.

$\frac{b-3}{2}-\frac{a-3}{2} = 10$

Multiply both sides by 2.

$(b-3) - (a-3) = 20$

Re-write without using parentheses.

$b-3-a+3 = 20$

Simplify.

$b-a = 20$

We need to find $b - a$ for the problem, so the answer is $\boxed{20\ \textbf{(D)}}$


See also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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